Divisor Count of 99

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Example of Use of Divisor Count Function

$\map {\sigma_0} {99} = 6$

where $\sigma_0$ denotes the divisor count function.


Proof

From Divisor Count Function from Prime Decomposition:

$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$

where:

$r$ denotes the number of distinct prime factors in the prime decomposition of $n$
$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$.


We have that:

$99 = 3^2 \times 11$

Thus:

\(\ds \map {\sigma_0} {99}\) \(=\) \(\ds \map {\sigma_0} {3^2 \times 11^1}\)
\(\ds \) \(=\) \(\ds \paren {2 + 1} \paren {1 + 1}\)
\(\ds \) \(=\) \(\ds 6\)


The divisors of $99$ can be enumerated as:

$1, 3, 9, 11, 33, 99$

This sequence is A018282 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).

$\blacksquare$